Can someone please help me derivate this

$\displaystyle h(x)=(x+1)^2(2x+1)^3 $

Thank you!

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- Dec 1st 2007, 06:38 AMCoachDerivation
Can someone please help me derivate this

$\displaystyle h(x)=(x+1)^2(2x+1)^3 $

Thank you! - Dec 1st 2007, 07:10 AMKrizalid
- Dec 1st 2007, 07:34 AMCoach
Thank you for your reply!

I do know the product rule in case you are talking about the uv´+vu´ rule, where the accent indicates derivated(I am sorry, I am very bad at these terms, since the textbook we are using is in Finnish). However, whenever I try to apply this rule, I get the wrong answer. Here is what I did.

$\displaystyle (X+1)^3*2(2X-1)^1+ (2X-1)^2*3(X+1)^2 $ - Dec 1st 2007, 07:50 AMKrizalid
Yes, that states the product rule. (Don't worry about the terms, I'm chilean and I have to learn them anyway.)

$\displaystyle \Big[ {f(x) \cdot g(x)} \Big]' = f'(x) \cdot g(x) + f(x) \cdot g'(x).$

The function is $\displaystyle h(x)=(x+1)^2(2x+1)^3,$

So $\displaystyle h'(x)=2(x+1)\cdot(2x+1)^3+3(x+1)^2(2x+1)^2\cdot2.$

Can you take it from there? - Dec 1st 2007, 07:58 AMCoach
Yes. Thank you!

But where did you get the last 2 from? - Dec 1st 2007, 08:03 AMKrizalid
It's the chain rule; the derivative of $\displaystyle (2x+1)^3$ is $\displaystyle 3(2x+1)^2\cdot(2x+1)'.$ That's why appears the $\displaystyle 2.$